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Sleeping Beauty Paradox

Q: Sleeping Beauty (SB) has volunteered for a test. The test is structured as follows. On a Sunday SB is put to sleep and after that a coin is tossed. If the coin

Comes up HEADS, SB is awoken on the following day (Monday) and asked her belief in the probability that the toss came up heads. The experiment ends.

Comes up TAILS, same as above except that at the end of the question her memory of the waking event and question is wiped out and she is awoken the next day (Tuesday) and asked the same question. The experiment ends.

The question is, when SB is awoken, what is SB's belief in the probability that it was heads. Note: At any given time, SB has no way to tell if it is a Monday or a Tuesday.

A: While it is tempting (and there are some arguments out there) to say the probability is \(\frac{1}{2}\), the reality isn't so. Here is why. Take a look at the following table.

The "X" indicates the situations when SB wakes up. Given that SB does not know which day of the week it is, she would guess \( P(Heads) = \frac{1}{3}\). If she knew that the day was a Monday, then we need only consider the Monday column and \(P(Heads) = \frac{1}{2}\). Likewise, if she knew it was a Tuesday then \(P(Tails) = 1\).

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Comments

While the answer is indeed 1/3, your argument is (barely) not right. It requires one more trivial step. But the halfers, those who favor the 1/2 answer, will simply dismiss it citing that step as the reason. They usually dismiss it with that step as well, but they never say why. Because they can't.

Instead of finding a flaw in the more explicit thirder reasoning, halfers claim that SB has no "new information" on which to base a change in the probability. But they never define what actually is necessary to do that, and it turns out you do have it.

Anyway, the original argument is that IF you are told it is Monday, you should have equal confidence that the coin landed heads or tails. This means that P(Heads&Monday)=P(Tails&Monday). Similarly, IF you are told that the coin landed tails, you should have equal confidence that it is Monday or Tuesday. This means that P(Tails&Monday)=P(Tails&Tuesday). Since I've listed the three exclusive possibilities, and they must be equal, it follows that all equal to 1/3.

The argument that directly contradicts the halfers is that, AT ANY TIME DURING THE EXPERIMENT, there are four possible combinations of day and coin, not three. Each has a prior probability to apply to a random moment of 1/4. Since she is awake, SB knows that the combination Tuesday&Heads is ruled out, so the updated probability for each of the other three is (1/4)/(1/4+1/4+1/4)=1/3.

The random experiment described in the beginning of your post consists of just a coin flip and there are only two possible outcomes (Heads or Tails). The "Tails and Monday", "Tails and Tuesday" events you employ are actually the same event (the outcome Tails). Hence, assuming a fair coin, P(Heads and Monday)=P(Heads)=1/2, P(Tails and Monday)= P(Tails and Tuesday)= P(Tails)=1/2.

Please also consider the following variance of the Sleeping Beauty Problem:Instead of one person the experiment is conducted to 10^6 persons. All but one will be awoken only once, whereas a randomly selected individual will be awoken 10^12 times using some drug that erases the memory of previous awakenings. All participants are informed about this setup.Before they are put to sleep the probability that they are the selected one, as calculated by each individual, is 10^-6, hence, they consider it highly improbable to be awoken more than once. According to thirders’ arguments, once they wake up, they should believe that the probability of NOT being the selected one is about 10^-6, since if the experiment is repeated an infinite amount of times for each individual, for every 10^6 awakenings roughly 1 will be associated with not being selected. Thus, according to thirders, just because you wake up (which you now all along that is certainly going to happen), you should think that you are the selected one, which is completely irrational, leading to absurd conclusions by ignoring basic probability rules on independence and Bayesian updating.Another argument thirders usually employ, is the betting odds. Let’s assume that each time a person wakes up is offered the following bet: If they bet that they are the selected one and win they will gain 1$, whereas if they loose they will loose 1000$. They are also informed that in case they are indeed the selected one, although the bet will be offered at each awakening, only the first one is valid. According to thirders they should always take the bet (since they are almost certain that they are the selected one) with an expected gain of about 1$. According to halfers the expected loss is about 1000$ and they would never take the bet. I would very much like to bet a million dollars against 1 million thirders and gain 1 billion dollars (minus of course the 1001$ that I will not get from the selected one)

Halfers always find an (invalid) reason to dismiss variations that involve repetition or betting. So instead, I'm going to propose a different alternate version:

Q: Sleeping Beauty (SB) has volunteered for a test. The test is structured as follows. On a Sunday SB is put to sleep in a room that is painted red. Sometime before Tuesday, a (fair) coin is tossed.

1) On Monday she is wakened in the same room. ("Awoke" in all its forms is an intransitive verb. SB can't be "awoken," but she can be "wakened." Its frequent use in this problem demonstrates that the authors are being pedantic.) She is interviewed. Then her memory is erased.

2) On Tuesday, if the coin comes up Heads, she is moved to a room that is painted blue. If it comes up Tails, she is left in the same room. Regardless of the coin, she is interviewed again.

3) In each interview, she is asked her belief in the probability that the toss will/did come up heads.

She can give different answers, depending on the color of the room she is in. What are they?

Obviously, her answer should be 100% in a blue room. It isn't hard to show that it should be 33% in a red room. And it isn't hard to accept, either, because increasing the probability in the case she knows didn't happen must be offset with a decrease in the one she knows she is in.

What people find hard to accept is that if she is in a red room, the problem is identical to your original. They can't reconcile the fact that not being able to know what happens on Tuesday after Heads is irrelevant if all she knows is that it is not Tuesday after Heads.

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